1. PE=KE
PE=mgh=mg•s•sinα
KE=mv²/2
mg•s•sinα=mv²/2
v=sqrt(2g•s•sinα) =
=sqrt(2•9.8•1.8•sin22°)=3.64 m/s
v=at => a=v/t
s=at²/2= vt²/2t=vt/2
t=2s/v=2•1.8/3.64 =0.99 s
2.
t₁=2t=0.99•2=1.98 s
v₁=2s/t₁=2•1.8/1.98= 1.82 m/s
PE=KE+W(fr)
W(fr) = F(fr)s=μ•mg•cosα•s
mg•s•sinα= mv₁²/2 + μ•mg•cosα•s
μ=(g•s•sinα- v₁²/2)/ g•cosα•s=
=tanα - (v₁²/2g•cosα•s)=
=tan22°-(1.82²/2•9.8•cos22°•1.8)=0.3
A box of mass 5.0 kg starting from rest slides down a 1.8 m long ramp inclined at an angle of 22.0 deg. from the horizontal. If the ramp is frictionless, how long will it take the box, starting from rest, to reach the bottom of the ramp? if the frictionless ramp is replaced with a ramp having friction between the box and ramp. The time for the box to reach the bottom of the ramp is observed to double. Find the coefficient of kinetic friction between the box and the new ramp.
1 answer
1 answer